InterviewSolution
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InterviewSolution
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The ration of mass per cent of C and H of an organic compound `(C_(x)H_(y)O_(z)) "is"6:1`. If one molecule of the above compound `(C_(x)H_(Y)O_(z))` contains half as much oxygen as required to burn one molecule of compound `C_(x)H_(Y)` compleltely to `CO_(2) and H_(2)O`. The empirial formula of compound `C_(x)H_(y)O_(z)` is:A. `C_(2)H_(4)O`B. `C_(3)H_(4)O_(2)`C. `C_(2)H_(3)O_(3)`D. `C_(3)H_(6)O_(3)` |
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Answer» Correct Answer - C Ratio of mass prcentage of carbon and hydrogen in the gives organic compound = 6:1 `therefore" " (12x)/(y)=(6)/(1)" " i.e, " " 2x = y` Number of oxygen atom in the compound`C_(x)H_(y)O_(z) = z " "......(i)` Equation of combustion of the compound `C_(x)H_(y) ` is : `C_(x)H_(y) + (x +(x)/(y)) O_(2) to xCO_(2)(g) +(y)/(2) H_(2)O(l)` Number of oxgyen atoms required for conbustion of `C_(x)H_(y) = (2x+(y)/(2))" ".....(i)` From (i) and (ii) ` z = (1)/(2) (2x +(y)/(2))= (1)/(2) (2x + (2x)/(2)) = (3x)/(2)` ` therefore " " x : y: z` ` X: 22x ":(3x)/(2)` |
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